98. Other changes in the Terms of a Fraction. 1. Reduce to an equivalent fraction having 25 for its numerator, Solution. — Observing that the proposed numerator, 25, is 5 times the given numerator, 5, we multiply both terms of f by 5, which gives f = 45. 2. Reduce of to an equivalent fraction having 15 for a numerator. SOLUTION.- Observing that the proposed numerator, 15, is 15, or of the given numerator, 6, we multiply both terms off by , or by 24, which gives 6 15 7 173 (a.) In the fractions written below substitute an appropriate denominator in place of x. Solution. — Observing that the proposed denominator, 24, is 3 times the given denominator, 8, we multiply both terms of by 3, which gives of 10. Reduce to sixths. SOLUTION.—Observing that the proposed denominator, 6, is of the given 38 denominator, 8, we multiply both terms of by 4 8 6 3 5 Note. — The following form of solution is sometimes adopted instead of the preceding: Since 1 = {1, must equal of t, which is it 6 5 5 Since 1 = must equal 6' 8 g (b.) Reduce 11. f to twentieths. 17. i and to twelfths. 99. Reduction to a Common Denominator. (a.) Fractions have a COMMON DENOMINATOR when their denomipators are alike. ILLUSTRATIONS. — and have the common denominator, 9. f and do not have a common denominator. (b.) Fractions having different denominators can be reduced to equivalent fractions having a common denominator. This is called REDUCING THEM TO A COMMON DENOMINATOR, and is illustrated in the last six examples of the preceding article. (c.) In reducing fractions to a common denominator 1st. Reduce the compound and complex fractions, if any, to simple ones. 2d. Reduce the simple fractions to their lowest terms, except when to do it would increase the labor of reducing to a common denominator. 3d. Select a convenient number for the common denominator. Usually it will be best to select the least common multiple of the denominators. 4th. Multiply both terms of each fraction by the number which gives the common denominator as the denominator of the results ing fraction. 1. Reduce , $, tt, t}, and 18 to a common denominator. = SOLUTION. — The least common multiple of the given denominators, 4, 5, 12, 15, and 20, is 60. Hence, we select 60 as the common denominator. Then, since 60 = 15 times 4, we multiply both terins off by 15, which gives Since 60 12 times 5, we multiply both terms of f by 12, which, etc. Hence, 48, it 18. is NOTE. — Many select the product of the denominators for a common denominator. This method usually involves larger numbers than the preceding, as will be seen in the following 2ND SOLUTION. — The product of the given denominators is 72000, which we select for the common denominator. To obtain this, we multiplied 4, the denominator of the first fraction, by all the other denominators, i. e. by 5, 12, 15, and 20. Hence, we must multiply the numerator by the same numbers, which gives 59888. We multiplied the denominator of the second fraction by all the other denominators, i. e. by 4, 12, 15, and 20; hence, we must multiply the numerator by the same numbers, which gives $2888, etc. 8. f and 14: 9. ff and Ms. 10. 14 and 1 24 17 7 11. f, g, and {. 15. and 5 3 12 14 1 3 12. $, 4, and f. 16. and 24' 41 274 13. 14, ti, and ot. 17. and 16' 48' 24 .5 6 .005 14. 4, 37, and 15. 18. and 35 12 34 100. Addition and Subtraction of Fractions. (a.) Fractions, like other numbers, must be of the same denomination in order to be added or subtracted. (b.) To be of the same denomination, they must be fractions of the same unit, and also have a common denominator. ILLUSTRATIONS. and cannot be added in their present form, any more than can 5 shillings and 3 pence. 11 of a gallon and 14 of a quart cannot be added, in their present form, any more than can 11 gallons and 11 quarts. (c.) Hence, to add or subtract fractions, it is necessary to reduce compound and complex fractions to simple ones, and fractions of different units (as, i of a bushel and t of a quart) to fractions of the same unit, and then to reduce the fractions thus obtained to a common denominator. The addition or subtraction can then be made by adding or subtracting the numerators, as illustrated in 77. 5 1. What is the sum of 6 3 41 11 7 ? 12 18 5 5 2 ones, we have SOLUTION. — Reducing the compound and complex fractions to simple 4} of and Hence, the problem be8 64 3* comes + f + it + 1g. Selecting 72 as the common denominator, and reducing the fractions to seventy-seconds, gives 4 + 4 + 4 + 4 243 18 2. What is the sum of ms of £1 + 24 shillings? SOLUTION. – 45 of £1 54 s., to which adding the 21 shillings gives 54 8. + 23 s. = 78. i of a shilling 10 d. Hence, of £1 + 24 s. = 7 8. 10 d. 3. What is the value of 78411 - 23633 ? SOLUTION. Selecting 90 as the common denominator, 78413 - 23637 696 = 54775 78457 — 2366 ones. 90ths. 78414 = 784 55 236 69 Note. — The written work in the margin shows the resemblance of the above to compound subtraction. 23638 547 76 54735 (d.) Perform the following operations : 16. 10 7.2 .6 5. +$. 11. 1-16 17. 8 .72 .25 .075 6. +3. 12. - be 18. .25 .2 .02 7..+. 13. 43 + 37. 19. .03 .3• 4.5 3.75 8. 1-3 14. 54 + 25 20. 7.5 27 1.3 9. + 15. 21. + 93 4.5 23 22. ftit f t . 27. +1 -1. 23. 11 + 24 +31 28. 21 +1-la. 24. 54 +31 + 2 + 1's. 29. ti + .7 + 35. 25. 15 + + 2 + 13. 30..4} +44 +.3. 26. j of zi – 4. 32. 1 - 4-4----3-74 33. 1-1-1-2-ois - do 34. 243 + 154 + 173-65 — 33 +6,15. 11 16 81 23 35. of of of 12 27 + 88 15 86.797 ++ 387. 5 .9 + .083 23 38. of a £ + $of a shilling. 39. i of a mile + $ of a yard. 40. of a gallon + of a quart. 41. of a lb. + g of an oz. + of à dwt. 42. / of a Ib of an 3 + of a 3. 43. z of a bushel + tá of a pk. it of a qt. ay of a pt. 2 24 2.5 17.57 44. of an acre 7 of a sq. rd. + 25 43 4.6 31. Ő + 3 13} 8q. yd. sq. in. |